(a) Let f(x)=cosx, then f′(x)=−sinx. In interval (0,
π
2
),f′(x)<0 Therefore, f(x) is strictly decreasing on (0,
π
2
) (b) Let f(x)=cos2x⇒f′(x)=−2sinx2x In interval (0,
π
2
),f′(x)<0 Because sin2x will either lie in the first or second quadrant which will give a positive value. Therefore, f(x) is strictly decreasing on (0,
π
2
) (c) Let f′(x)=cos3x ⇒f′(x)=−3sin3x. In Interval (0,
π
3
),f′(x)<0 Because sin3x will either lie in the first or second quadrant which will give a positive value. Therefore, f(x) is strictly decreasing on (0,
π
3
). When x∈(
π
3
,
π
2
), then f′(x)>0 Because sin3x will lie in the third quadrant. Therefore, f(x) is not strictly decreasing on (0,
π
2
) (d) Let f(x)=tanx⇒f′(x)=sec2x. In Interval x∈(0,
π
2
),f′(x)>0 Therefore, f(x) is not strictly decreasing on (0,